Question

已知函数f(x)=2sinxcosx+cos2x

已知函数f(x)=2sinxcosx+cos2x(x属于R)
1.求f(x)的最小正周期和最大值
2.若θ为锐角，且f(θ+π/8)=3分之根号2，求tan2θ的值

Answer 1

f(x)=sin2x+cos2x
=√2sin(2x+π/4)
所以T=2π/2=π
最大值=√2

f(θ+π/8)=√2sin(2θ+π/4+π/4)
=√2cos2θ
=√2/3
cos2θ=1/3
θ锐角则sin2θ>0
sin²2θ+cos²2θ=1
sin2θ=2√2/3
tan2θ=2√2

Answer 2

x1>0,x2<0
|x1|=2|x2|
x1=-2x2

所以x1+x2=-x2
x1x2=-2x2^2
则(x1x2)/(-2)=(x1+x2)^2

x1+x2=m/2
x1x2=-5m/4-3
所以-5m/8+3/2=m^2/4
2m^2+5m-12=0
(2m-3)(m+4)=0

异号则x1x2=-5m/4-3<0
m>-12/5
所以m=3/2